A block of mass m tied to a string is lowered by a distance d, at a constant acceleration of `g//3`. The work done by the string is

Options

A. `+(3mgd)/(4)`
B. `-(mgd)/(4)`
C. `-(2)/(3)mgd`
D. `+mgd`

Answer

To solve the problem, we need to determine the work done by the string when a block of mass \( m \) is lowered by a distance \( d \) with a constant acceleration of \( \frac{g}{3} \).

### Step-by-Step Solution:

1. **Identify Forces Acting on the Block**:
- The weight of the block acting downwards: \( W = mg \)
- The tension in the string acting upwards: \( T \)
- The net force acting on the block is given by Newton's second law: \( F_{net} = ma \).

2. **Set Up the Equation of Motion**:
- The block is accelerating downwards with \( a = \frac{g}{3} \).
- According to Newton's second law, we can write:
\[ mg - T = ma \]
- Substituting \( a = \frac{g}{3} \):
\[ mg - T = m \left(\frac{g}{3}\right) \]

3. **Solve for Tension \( T \)**:
- Rearranging the equation gives:
\[ T = mg - m\left(\frac{g}{3}\right) \]
- Factor out \( mg \):
\[ T = mg \left(1 - \frac{1}{3}\right) = mg \left(\frac{2}{3}\right) \]
- Thus, the tension \( T \) is:
\[ T = \frac{2}{3}mg \]

4. **Calculate the Work Done by the String**:
- The work done by the tension in the string when the block is lowered by a distance \( d \) is given by:
\[ W = T \cdot d \cdot \cos(\theta) \]
- Here, the angle \( \theta \) between the tension (upward) and the displacement (downward) is \( 180^\circ \), so \( \cos(180^\circ) = -1 \):
\[ W = T \cdot d \cdot (-1) = -T \cdot d \]
- Substituting the value of \( T \):
\[ W = -\left(\frac{2}{3}mg\right) \cdot d \]
- Therefore, the work done by the string is:
\[ W = -\frac{2}{3}mgd \]

5. **Final Answer**:
- The work done by the string is:
\[ W = -\frac{2}{3}mgd \]